Exercise 7.  Find  ,   when  

7 (a).  .  

Solution 7 (a).

See text and/or instructor's solution manual.

Answer.   .  

Solution.  Factor the denominator and get

          ,

and conclude that    has simple zeros at  ,  ,  ,  and  .   

Use the techniques for determining the order of poles: apply  Corollaries 7.5 and 7.8  in Section 7.4 to conclude that

the integrand    

has simple poles at  ,  ,  ,  and  .   

                    The point    that lies inside the contour  .  

        Now apply Theorem 8.1 the Cauchy's Residue Theorem.

The point    is the only singularity that lie inside the contour  .  

Since   is a simple pole, by Theorem 8.2 the residue is calculated as follows:

                    

The "Residue Calculus" gives the value of the integral:

                                

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > limit( (z-I*3^(1/2))*1/(3*z^4+10*z^2+3), z=I*3^(1/2) );

               

We are really done.   

Aside.  Both and are capable of finding residues.

     > residue( 1/(3*z^4+10*z^2+3), z=I*3^(1/2) );

               

     > 2*Pi*I*residue(1/(3*z^4+10*z^2+3),z=I*3^(1/2));

               

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell